10-2 of , Rhombi, and Kites

Find the of each , , or .

3.

1. SOLUTION: SOLUTION:

ANSWER: ANSWER:

4. PEP RALLY Suki is designing posters for the Homecoming game. Her design is shown at the right. What is the area of the poster in feet?

2.

SOLUTION:

SOLUTION: ANSWER: The area A of a trapezoid is .

She will need square feet of fabric.

ANSWER: 4.375 ft²

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ALGEBRA Find x. 7. 5.

SOLUTION: SOLUTION:

ANSWER: 6.3 ft ANSWER: STRUCTURE Find the area of each trapezoid, 8 cm rhombus, or kite.

6.

8.

SOLUTION:

SOLUTION: h = 13, b1 = 18 and b2 = 24

ANSWER: 6.6 in. ANSWER:

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11. 9. SOLUTION: SOLUTION: d1 = 16 in. and d2 = 17 in. h = 23, b1 = 22 and b2 = 37

ANSWER:

ANSWER:

10. 12.

SOLUTION: SOLUTION: d = 7 cm. and d = 6 + 9 = 15 cm d1 = 22 and d2 = 24 1 2

ANSWER: ANSWER:

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15. heartleaf plant Refer to the image on Page 739.

SOLUTION: 13.

SOLUTION:

d1 = 11 ft. and d2 = 25 ft

ANSWER: 24.5 square microns

16. eye of a fly Refer to the image on Page 739.

SOLUTION: ANSWER: The figure consists of two trapezoids.

MICROSCOPES Find the area of the identified portion of each magnified image. Assume that the identified portion is either a trapezoid, rhombus, or kite. Measures are provided in microns. 14. human skin Refer to the image on Page 739. ANSWER:

9.9 square microns SOLUTION:

ANSWER: 26 square microns

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17. JOBS Jimmy works on his neighbors’ yards after 19. The area of a rhombus is 168 square centimeters. If school to earn extra money to buy a car. He is going one is three times as long as the other, what to plant grass seed in Mr. Troyer’s yard. What is the are the lengths of the ? area of the yard? SOLUTION: The area A of a rhombus is one half the product of the lengths of its diagonals, d1 and d2.

SOLUTION:

ANSWER:

Therefore, the diagonals are of length 10.6 cm. and ALGEBRA Find each missing length. 31.7 cm. 18. One diagonal of a kite is twice as long as the other diagonal. If the area of the kite is 240 square inches, ANSWER: what are the lengths of the diagonals? 10.6 cm, 31.7 cm

SOLUTION: 20. A trapezoid has lengths of 12 and 14 feet with The area A of a kite is one half the product of the an area of 322 square feet. What is the height of the trapezoid? lengths of its diagonals, d1 and d2. SOLUTION: The area A of a trapezoid is one half the product of the height h and the sum of the lengths of its bases, b1 and b2.

Therefore, the diagonals are of length 15.5 in. and 31.0 in.

ANSWER: ANSWER: 15.5 in., 31.0 in. 24.8 ft

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21. A trapezoid has a height of 8 meters, a base length of 22. HONORS Estella has been asked to join an honor 12 meters, and an area of 64 square meters. What is society at school. Before the first meeting, new the length of the other base? members are asked to sand and stain the front side of a piece of wood in the of an isosceles SOLUTION: trapezoid. What is the surface area that Allison will The area A of a trapezoid is one half the product of need to sand and stain? the height h and the sum of the lengths of its bases, b1 and b2.

SOLUTION: ANSWER: The required area is the difference between the 4 m larger trapezoid and the smaller trapezoid. The area A of a trapezoid is one half the product of the height h and the sum of the lengths of its bases, b1 and b2.

ANSWER:

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For each figure, provide a justification showing that .

24.

23. SOLUTION: Set the area of the rhombus equal to the sum of the SOLUTION: areas of the two with bases d and d . Set the area of the kite equal to the sum of the areas 1 2 ZWX = ZYX by SSS of the two triangles with bases d and d . 1 2

ANSWER: ANSWER: The area of and the area of The area of and the area of

. Therefore, the area of . Therefore, the area of

, and the area of . , and the area of . The area of kite FGHJ is equal to the area of The area of rhombus WXYZ is equal to the area of the area of . After the area of . After simplification, the area of kite FGHJ is equal to simplification, the area of rhombus WXYZ is equal to . .

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25. CRAFTS Ashanti is competing in a kite festival. The SENSE-MAKING Find the area of each yellow, red, orange, green, and blue pieces of her kite with the given vertices. design shown are all congruent rhombi. 26. A(–8, 6), B(–5, 8), C(–2, 6), and D(–5, 0) a. How much fabric of each color does she need to SOLUTION: buy? Graph the quadrilateral. b. Competition rules require that the total area of each kite be no greater than 200 square inches. Does Ashanti’s kite meet this requirement? Explain.

The quadrilateral is a kite. The area A of a kite is one SOLUTION: half the product of the lengths of its diagonals, d1 and The area of the yellow rhombus is or 24 in2. d2. The lengths of the diagonals are 6 units and 8 units. Therefore, the area of the kite is Since the yellow, red , orange, green, and blue pieces area all congruent rhombi, each have an area of 24 in 2. ANSWER: The area of the purple kite-shaped piece is 24 units² or 20 in2. The total area of the entire kite is 24(5) + 20 or 140 in 2, which is less than the maximum 200 in2 allowed. Therefore, her kite meets this requirement.

ANSWER: a. each of yellow, red, orange, green, and blue; of purple b. Yes; her kite has an area of , which is less than 200 in2.

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27. W(3, 0), X(0, 3), Y(–3, 0), and Z(0, –3) 28. METALS When magnified in very powerful microscopes, some metals are composed of grains SOLUTION: that have various polygonal . Graph the quadrilateral.

a. What is the area of figure 1 if the grain has a height of 4 microns and bases with lengths of 5 and 6 microns? The quadrilateral is a rhombus. The area A of a b. If figure 2 has diagonal lengths of rhombus is one half the product of the lengths of its 3.8 microns and 4.9 microns, what is the area of the diagonals, d and d . 1 2 grain? The lengths of the diagonals are 6 units each. Therefore, the area of the kite is SOLUTION: a. Figure 1 is a trapezoid.

ANSWER: 18 units² The area of figure 1 is 22 square microns.

b. Figure 2 is a rhombus.

The area of figure 2 is about 9.3 square microns.

ANSWER: a. 22 square microns b. 9.3 square microns

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29. PROOF The figure at the right is a trapezoid that nearest tenth, if necessary. consists of two congruent right triangles and an isosceles . In 1876, James A. Garfield, the 20th president of the United States, discovered a proof of the using this diagram. Prove that . 30.

SOLUTION: Both diagonals are perpendicular bisectors, so the figure is a rhombus and all four triangles are congruent. All of the sides are 12 feet, so the is 48 feet. SOLUTION: Following is an algebraic proof. Use trigonometry to find the lengths of the diagonals.

The trapezoid is on its side, so the bases are x and y and the height is x + y. Set the area of the trapezoid equal to the sum of the areas of the triangles and simplify.

ANSWER: The area of a trapezoid is So,

. Now find the area.

The area of . The area of . Set the area of the trapezoid equal to the combined areas of the triangles to get . Multiply

by 2 on each side: . When simplified, . ANSWER: DIMENSIONAL ANALYSIS Find the perimeter 48 ft; 129.4 ft2 and area of each figure in feet. Round to the eSolutions Manual - Powered by Cognero Page 10 10-2 Areas of Trapezoids, Rhombi, and Kites

32. 31. SOLUTION: SOLUTION: The figure is a kite because one of the diagonals is a Use the 30-60-90 triangle to find the of perpendicular bisector. Find the perimeter. Convert the . The base of the triangle is the units to feet. 0.5(12 – 8) = 2.

Don't forget to use dimensional analysis to convert the units to feet.

Use the 45-45-90 triangle to find the lengths of the congruent parts of the diagonals.

d1 = 3 + 3 = 6 ANSWER: 2.3 ft; 0.24 ft2 Use the Pythagorean theorem to find the other piece of d2.

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What is the significance of this value?

SOLUTION: a. When x = 2, the long diagonal will be in pieces of length 2 and 10.

Now find the area of the kite. b. When x = 4, the long diagonal will be in pieces of length 4 and 8.

When x = 6, the long diagonal will be in pieces of ANSWER: length 6 and 6. 67.6 ft; 267.8 ft2

33. MULTIPLE REPRESENTATIONS In this problem, you will investigate of kites.

When x = 8, the long diagonal will be in pieces of length 8 and 4. a. GEOMETRIC Draw a kite like the one shown if x = 2. b. GEOMETRIC Repeat the process in part a for three x-values between 2 and 10 and for an x-value of 10. The overall length of the kite should remain 12 cm. c. TABULAR Measure and record in a table the perimeter of each kite, along with the x-value. When x = 10, the long diagonal will be in pieces of d. GRAPHICAL Graph the perimeter versus the x- length 10 and 2. value using the data from your table. e. ANALYTICAL Make a conjecture about the value of x that will minimize the perimeter of the kite. eSolutions Manual - Powered by Cognero Page 12 10-2 Areas of Trapezoids, Rhombi, and Kites

hypotenuse of a with legs of 2 cm and 4 cm. Thus, the length of two legs = and the length of the other two sides = . So, the perimeter of the kite is cm.

For x = 10: Two sides of the kite are the length of the c. Use the Pythagorean Theorem to first find the hypotenuse of right triangle with legs of 2 cm and 10 lengths of the sides of the kite and then find its cm and the other two sides are the length of the perimeter. hypotenuse of a right triangle with legs of 2 cm each.

For x = 2: Two sides of the kite are the length of the Thus, the length of two legs = hypotenuse of right triangle with legs of 2 cm each and the length of the other two sides = and the other two sides are the length of the . So, the perimeter of the kite is hypotenuse of a right triangle with legs of 2 cm and cm. 10 cm. Thus, the length of two legs = and the length of the other two sides x P 2 cm 26.1 = . So, the perimeter of the kite 4 cm 25.4 is cm. 6 cm 25.3

8 cm 25.4 For x = 4: Two sides of the kite are the length of the 10 cm 26.1 hypotenuse of right triangle with legs of 2 cm and 4 cm and the other two sides are the length of the hypotenuse of a right triangle with legs of 2 cm and 8 d. cm. Thus, the length of two legs = and the length of the other two sides = . So, the perimeter of the kite is cm.

For x = 6: Two sides of the kite are the length of the hypotenuse of right triangle with legs of 2 cm and 6 cm and the other two sides are the length of the hypotenuse of a right triangle with legs of 2 cm and 6 cm. Thus, the length of two legs = and the length of the other two sides = e. Based on the graph, the perimeter will be minimized when x = 6. This value is significant . So, the perimeter of the kite is because when x = 6, the figure is a rhombus. cm. ANSWER: For x = 8: Two sides of the kite are the length of the a. hypotenuse of right triangle with legs of 2 cm and 8 cm and the other two sides are the length of the eSolutions Manual - Powered by Cognero Page 13 10-2 Areas of Trapezoids, Rhombi, and Kites

d.

b.

e. Sample answer: Based on the graph, the perimeter will be minimized when x = 6. This value is significant because when x = 6, the figure is a rhombus.

c.

x P 2 cm 26.1 4 cm 25.4 6 cm 25.3 8 cm 25.4 10 cm 26.1 eSolutions Manual - Powered by Cognero Page 14 10-2 Areas of Trapezoids, Rhombi, and Kites

34. CRITIQUE ARGUMENTS Antonio and Madeline want to draw a trapezoid that has a height of 4 units and an area of 18 square units. Antonio says that only one trapezoid will meet the criteria. Madeline disagrees and thinks that she can draw several different trapezoids with a height of 4 units and an area of 18 square units. Is either of them correct? We have two variables, x and a. They are a part of Explain your reasoning. two right triangles, so we can use the Pythagorean Theorem for each triangle and then combine the SOLUTION: equations. There is more than one trapezoid with a height of 4 units and an area of 18 square units. The sum of the bases of the trapezoid has to be 9, so one possibility is a trapezoid with bases of 4 and 5 units and a height of 4 units. Another is a trapezoid with bases of 3 and 6 units and a height of 4 units. Therefore, Madeline is correct.

We now have x2 on the left side of each equation. Use substitution.

ANSWER: Madeline; sample answer: There is more than one trapezoid with a height of 4 units and an area of 18 square units. The sum of the bases of the trapezoid has to be 9, so one possibility is a trapezoid with bases of 4 and 5 units and a height of 4 units. Use the value of a to find the value of x. Another is a trapezoid with bases of 3 and 6 units and a height of 4 units.

35. CHALLENGE Find x in ABCD.

ANSWER: 7.2

SOLUTION: Let the two segments of the side be a and 15 – a.

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36. OPEN-ENDED Draw a kite and a rhombus with an 37. REASONING If the areas of two rhombi are equal, area of 6 square inches. Label and justify your are the perimeters sometimes, always, or never drawings. equal? Explain.

SOLUTION: SOLUTION: If the areas are equal, it means that the products of the diagonals are equal. The only time that the perimeters will be equal is when the diagonals are also equal, or when the two rhombi are congruent. Therefore, the statement is sometimes true.

Since the area formula for both a rhombus and a kite For example, is one half the product of the lengths of the two When the diagonals are 6 and 8, the area is 24 and diagonals, if the area is 6 square inches, the product the perimeter is 26. of the two diagonals must be 12. I used 3 and 4 When the diagonals are 12 and 4, the area is 24 and inches for the diagonals of the rhombus and 2 and 6 the perimeter is 32. inches for the diagonals of the kite. ANSWER: Both diagonals must be perpendicular bisectors for Sometimes; sample answer: If the areas are equal, it the rhombus and one diagonal must be a means that the products of the diagonals are equal. perpendicular bisector for the kite. The only time that the perimeters will be equal is when the diagonals are also equal, or when the two ANSWER: rhombi are congruent.

Sample answer: Since the area formula for both a rhombus and a kite is one half the product of the lengths of the two diagonals, if the area is 6 square inches, the product of the two diagonals must be 12. I used 3 and 4 inches for the diagonals of the rhombus and 2 and 6 inches for the diagonals of the kite.

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38. WRITING IN MATH How can you use 39. The Jays’ team logo is a rhombus, as shown. trigonometry to find the area of a figure?

SOLUTION: Trigonometry can be used with known and sides to find unknown sides. For example;

The team’s manager makes an enlargement of the logo in which each is 4 times larger than the version shown here. Which of the following best describes how the areas of the logos compare?

A The area of the enlargement is 2 times the area of The height of the parallelogram can be found using the original. the sine function: B The area of the enlargement is 4 times the area of the original. C The area of the enlargement is 8 times the area of the original. D The area of the enlargement is 16 times the area And the area can be found as usual: of the original. SOLUTION: Multiply the diagonals by 4 and compare the areas of the original and enlargement.

Area of original = (4.6)(1.5) = 6.9 sq. cm

ANSWER: Area of enlargement = 4(4.6)(4(1.5)) = 110.4 sq. cm

Sample answer: You can use trigonometry and known 16(6.9 sq. cm) = 110.4 sq. cm and side measures to find unknown triangular

measures that are required to calculate the area. The area of the enlargement is 16 times the area of the original.

ANSWER: D

40. Li Mei drew the trapezoid shown below as the plan for a new concrete driveway apron. The contractor will charge $9.50 per square foot to install the concrete. The homeowner will seal the concrete with sealer that costs $40 for a container that covers 80 square feet.

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Which of the following statements about the 41. One diagonal of a rhombus is four times as long as driveway apron are true? Select all that apply. the other diagonal. If the area of the rhombus is 288 square meters, what are the lengths of the diagonals? A It will cost $1216 for the concrete. B The homeowner will need four containers of sealer SOLUTION: to seal the concrete. Use the formula for the area of a rhombus to solve. C A $1300 budget for the concrete and sealer is enough for the project. D The area of concrete is 288 square feet.

SOLUTION: The area of Li Mei's trapezoid is:

The cost of the concrete is $9.50 per square foot. Multiply that by the area. The two diagonals are 12 meters and 48 meters.

128(9.5) = $1,216 ANSWER: 12 meters and 48 meters The cost of the sealer is $40 to cover 80 square feet, so the homeowner will need 2 containers. 42. The figure shows a kite that Alex created using geometry software. The total cost is $1,216 + 2(40) = $1,296

It will cost $1216 for the concrete, so choice A is true. The homeowner will need two not four containers of sealer to seal the concrete, so choice B is false. A $1300 budget for the concrete and sealer is enough for the project, since the total cost is $1,296 so choice C is true. The area of concrete is 128 square feet, not 288 Which of the following will result in a kite with twice square feet, so choice D is false. the area of ABCD? ANSWER: I. Double the length of each diagonal. A, C II. Double the length of . III. Double the length of .

A I only B II only C III only D II and III only

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The area of kite ABCD is: 43. The sides of a soccer goal are in the form of a trapezoid. The bottom base of the trapezoid is 6.5 feet long, the top base of the trapezoid is 4.25 feet long, and the height of the trapezoid is 8 feet high. Find the area of one side of the soccer goal.

A 38 square feet If you double the length of each diagonal (option I), B 40 square feet you get: C 43 square feet D 50 square feet

SOLUTION: Use the formula for the area of a trapezoid. This is not double the area of 3.5, so option I is not correct. 6.5 feet long, the top base of the trapezoid is 4.25 feet long, and the height of the trapezoid is 8 feet high. If you double the length of diagonal (option II), you get:

This is double the area of 3.5, so Area is 43 square feet. The correct choice is C. option II is correct. ANSWER: If you double the length of diagonal (option III), C you get:

This is double the area of 3.5, so option III is also correct.

The correct choice is D, as II and III only are correct.

ANSWER: D

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44. MULTI-STEP The area of a kite is 50 square inches.

a. If each diagonal is a whole number, how many distinct diagonal pairs have whole number lengths? b. Which diagonal length, in inches, will guarantee that the kite is also a rhombus? c. If the area of the kite is doubled, how many distinct diagonal pairs have whole number lengths?

SOLUTION: a. If each diagonal is a whole number, the diagonal pairs that have whole number lengths must have a product of 100, because the formula for area of a kite is half the product of the diagonals.

The whole number lengths for diagonals are 1 and 100, 2 and 50, 4 and 25, 5 and 20, and 10 and 10. There are 5 possible pairs.

b. Which diagonal length, in inches, will guarantee that the kite is also a rhombus? If the diagonals are 10 inches long, the figure is a rhombus.

c. If the area of the kite is doubled to 100 square inches, the whole number lengths for diagonals are 1 and 200, 2 and 100, 4 and 50, 5 and 40, 8 and 25, and 10 and 10. There are 6 possible pairs.

ANSWER: a. 5 b. 10 c. 6

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